In the circuit shown (Fig. 19-8), find (a) the total inductance and (b) the time constant. Assume that a current of 10 A is flowing when the switch S is opened. Find (c) the current 2 s later.
(a) Reduce the series-parallel combination .of inductances to its equivalent.
Parallel: L_a=\frac{L_1 L_2}{L_1+L_2}=\frac{5(1)}{5+1}=\frac{5}{6}=0.83\ H
Parallel: L_b=\frac{L_3 L_4}{L_3+L_4}=\frac{4(4)}{4+4}=\frac{16}{8}=2.0\ H
Then series: L_T=L_a+L_b=0.83+2.0=2.83\ H
(b) The time constant of the circuit is the ratio of total inductance to the total resistance of the circuit.
T=\frac{L_T}{R}=\frac{2.83}{5}=0.57\ s(c) Write the formula for decaying current. The V/R value is 10 A.
i=\frac{V}{R} e^{-R t / L} (19-7)
\begin{aligned}-\frac{R t}{L_T} & =-\frac{5(2)}{2.83}=-3.5 \\ e^{-3.5} & =0.030\end{aligned}Then i = 10(0.030) = 0.30 A